I'm looking for a solution of
\begin{equation}
u_t(x,t)+c\ u_x(x,t)+\gamma\ u(x,t) = 0,
\end{equation}
for $x,c,\gamma\in\mathbb{R}$ and $t>0$ and $u(x,0)=u_0(x)$. The exercise is to solve it with Fourier transform.
My idea was using fourier transform in $x$ so I get
\begin{equation}
\hat{u}_t(\xi,t)+i \ c\ \xi\ \hat{u}(\xi,t)+\gamma \hat{u}(\xi,t) = 0, \text{or}\\
\hat{u}_t(\xi,t)+ (i \ c\ \xi\ + \gamma) \hat{u}(\xi,t) = 0
\end{equation}
Solving the ode gives
\begin{equation}
\hat{u}(\xi,t)= A\exp{\left(-(i\ c\ \xi + \gamma)t\right)}
\end{equation}
Now using inverse fourier transform gives
\begin{align}
u(x,t) &= A \int_{-\infty}^{\infty} \exp{\left(-(i\ c\ \xi + \gamma)t\right)} \exp(i\ x \xi)\ d\xi \\
&= A \exp(-\gamma t) \int_{-\infty}^{\infty} \exp(i \xi (x-c))\ d\xi
\end{align}
which results in Dirac Delta function?! Is the ansatz wrong or is there a mistake somewhere?
Thank you!
Best Answer
Solving the ODE in Fourier space gives $$\hat u(\xi, t) = \hat u_0(\xi) \exp(-(ic\xi + \gamma)t)$$ where $\hat u_0$ is the Fourier transform of $u(x,0) = u_0(x)$. Then taking the inverse Fourier transform gives $$\begin{align*}u(x,t) &= \int_{-\infty}^\infty \hat u_0(\xi) \exp(-(ic\xi + \gamma)t) \exp(ix\xi) \, d\xi \\ &= \exp(-\gamma t)\int_{-\infty}^\infty \hat u_0(\xi)\exp(i\xi(x-ct)) \, d\xi \\ &= \exp(-\gamma t) u_0(x-ct)\end{align*} $$ where the last line follows from the integral simply being the inverse Fourier transform of $u_0$ evaluated at $x-ct$. You can plug this function into the PDE and see that it is in fact a solution.